CFM of Cool Air as a Function of Heat Load 1

[cjuwilkins]

Hello,

I wrote a lengthier post but I don't think my question deserves a large amount of text. If you need more information please ask for it, I'm trying to make my question as quick and as simple as possible.

I am trying to find out how to relate the amount of cold air "in" I need for a heat load (I am not in mechanical, I am trying to say I have the amount of heat, in btu/hr, given off by some mechanical equipment). So all the heating in this room is provided by the mechanical equipment, and all of the cooling will be provided by fresh air from the outside, through some ducts (I am trying to choose the fans for these ducts). I have average temperature during different months of the year. Can I just use this formula, rearragned for "q" (volume flow rate), and subbing the heat load from the mech equipments as h_s? And then dt would be (T_outside - T_inside)?

hs = 1.08*q*dt

If not, what is the relationship? I feel like there must be one, it'd be enough from someone here to provide me with a reference to read, as I am having trouble googling for something that can help.

Thanks in advance,

cjuwilkins

 

[tys90]

To keep in the theme of keeping posts brief, if you assume all that heat load is sensible, you're correct in your reasoning. Is there a roof or and exterior walls to this room? You will need to account for heat transfer through them if there are.

 

[IRstuff]

Also, what is the actual cooling or temperature requirement? Just because you move x CFM of air, doesn't mean that it'll all get mixed and heated as desired.

At some level, starting with outside air only is a recipe non-starter, particularly if there are people in the room. The amount of flow required might be excessive for humans in the room, or the humidity of the outside air is so high that the "effective" cooling is nowhere near what the occupants are actually comfortable with.

TTFN

FAQ731-376

 

[cjuwilkins]

Thanks for the advice guys. To IRstuff, the room is pretty small and the exhaust and intake vents will be in opposite corners of the room. I am fairly confident we'll be satisfied, assuming I choose the right fan speed.

Just checking, are there any other formulas that might be useful to me? The amount of flow is indeed pretty "excessive", and makes me question how reliable this equation is. It is asking for ~1800 CFM in the winter months, when the air will be around
-10 C (14 F). The room is around 35 C (95 F) right now, and we want it be 25 C (77 F).

 

[cjuwilkins]

EDIT: Maybe I should mention that for dT, I used 95F - 14F.

 

[tys90]

If you want it to be 77F then you need to use 77F instead of 95. 1,800 CFM at those temperatures would indicate a load of 122,000 BTUs which is a LOT. 10 tons of cooling for one "small" room is data center type loads. Make sure you are using correct units. h_s is in BTU/hr, q is in CFM and temperature is in F

 

[cjuwilkins]

I really appreciate your advice. I'll look into whether the different numbers I used are reasonable, and at least now I know how to use the equation. Thanks again!

 

[IRstuff]

The A/C salemen use a rule of thumb between 500 and 700 sq ft per ton of A/C.

Perhaps it's time you revealed what your heat load is.

1800 CFM for a 14x14 room would be on the order 55 changes per hour, which does seem a bit extreme. Assuming a 1 ft square register, that would give you an outlet air speed of 20 mph.

You still have not indicated where the heat load is located in the room. If it's against the walls, the air flow will probably not get to the heat source, since the walls, edges, and corners are stagnant.

A more germane question is that if this is a piece of equipment, why not duct the majority airflow directly to the equipment? This would ensure that the air gets to where it needs to go, and ensures that people won't get a massive air blast on them while they're in the room.

TTFN

FAQ731-376